3.200 \(\int \frac{(a+b x^3)^{3/2} (A+B x^3)}{x^4} \, dx\)

Optimal. Leaf size=110 \[ \frac{\left (a+b x^3\right )^{3/2} (2 a B+3 A b)}{9 a}+\frac{1}{3} \sqrt{a+b x^3} (2 a B+3 A b)-\frac{1}{3} \sqrt{a} (2 a B+3 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )-\frac{A \left (a+b x^3\right )^{5/2}}{3 a x^3} \]

[Out]

((3*A*b + 2*a*B)*Sqrt[a + b*x^3])/3 + ((3*A*b + 2*a*B)*(a + b*x^3)^(3/2))/(9*a) - (A*(a + b*x^3)^(5/2))/(3*a*x
^3) - (Sqrt[a]*(3*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/3

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Rubi [A]  time = 0.0840313, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {446, 78, 50, 63, 208} \[ \frac{\left (a+b x^3\right )^{3/2} (2 a B+3 A b)}{9 a}+\frac{1}{3} \sqrt{a+b x^3} (2 a B+3 A b)-\frac{1}{3} \sqrt{a} (2 a B+3 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )-\frac{A \left (a+b x^3\right )^{5/2}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)^(3/2)*(A + B*x^3))/x^4,x]

[Out]

((3*A*b + 2*a*B)*Sqrt[a + b*x^3])/3 + ((3*A*b + 2*a*B)*(a + b*x^3)^(3/2))/(9*a) - (A*(a + b*x^3)^(5/2))/(3*a*x
^3) - (Sqrt[a]*(3*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/3

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{x^4} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2} (A+B x)}{x^2} \, dx,x,x^3\right )\\ &=-\frac{A \left (a+b x^3\right )^{5/2}}{3 a x^3}+\frac{\left (\frac{3 A b}{2}+a B\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,x^3\right )}{3 a}\\ &=\frac{(3 A b+2 a B) \left (a+b x^3\right )^{3/2}}{9 a}-\frac{A \left (a+b x^3\right )^{5/2}}{3 a x^3}+\frac{1}{6} (3 A b+2 a B) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^3\right )\\ &=\frac{1}{3} (3 A b+2 a B) \sqrt{a+b x^3}+\frac{(3 A b+2 a B) \left (a+b x^3\right )^{3/2}}{9 a}-\frac{A \left (a+b x^3\right )^{5/2}}{3 a x^3}+\frac{1}{6} (a (3 A b+2 a B)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^3\right )\\ &=\frac{1}{3} (3 A b+2 a B) \sqrt{a+b x^3}+\frac{(3 A b+2 a B) \left (a+b x^3\right )^{3/2}}{9 a}-\frac{A \left (a+b x^3\right )^{5/2}}{3 a x^3}+\frac{(a (3 A b+2 a B)) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^3}\right )}{3 b}\\ &=\frac{1}{3} (3 A b+2 a B) \sqrt{a+b x^3}+\frac{(3 A b+2 a B) \left (a+b x^3\right )^{3/2}}{9 a}-\frac{A \left (a+b x^3\right )^{5/2}}{3 a x^3}-\frac{1}{3} \sqrt{a} (3 A b+2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0473543, size = 80, normalized size = 0.73 \[ \frac{1}{9} \left (\frac{\sqrt{a+b x^3} \left (-3 a A+8 a B x^3+6 A b x^3+2 b B x^6\right )}{x^3}-3 \sqrt{a} (2 a B+3 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)^(3/2)*(A + B*x^3))/x^4,x]

[Out]

((Sqrt[a + b*x^3]*(-3*a*A + 6*A*b*x^3 + 8*a*B*x^3 + 2*b*B*x^6))/x^3 - 3*Sqrt[a]*(3*A*b + 2*a*B)*ArcTanh[Sqrt[a
+ b*x^3]/Sqrt[a]])/9

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Maple [A]  time = 0.02, size = 101, normalized size = 0.9 \begin{align*} A \left ( -{\frac{a}{3\,{x}^{3}}\sqrt{b{x}^{3}+a}}+{\frac{2\,b}{3}\sqrt{b{x}^{3}+a}}-\sqrt{a}b{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ) \right ) +B \left ({\frac{2\,b{x}^{3}}{9}\sqrt{b{x}^{3}+a}}+{\frac{8\,a}{9}\sqrt{b{x}^{3}+a}}-{\frac{2}{3}{a}^{{\frac{3}{2}}}{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(3/2)*(B*x^3+A)/x^4,x)

[Out]

A*(-1/3*a*(b*x^3+a)^(1/2)/x^3+2/3*b*(b*x^3+a)^(1/2)-a^(1/2)*b*arctanh((b*x^3+a)^(1/2)/a^(1/2)))+B*(2/9*b*x^3*(
b*x^3+a)^(1/2)+8/9*a*(b*x^3+a)^(1/2)-2/3*a^(3/2)*arctanh((b*x^3+a)^(1/2)/a^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/2)*(B*x^3+A)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.81138, size = 405, normalized size = 3.68 \begin{align*} \left [\frac{3 \,{\left (2 \, B a + 3 \, A b\right )} \sqrt{a} x^{3} \log \left (\frac{b x^{3} - 2 \, \sqrt{b x^{3} + a} \sqrt{a} + 2 \, a}{x^{3}}\right ) + 2 \,{\left (2 \, B b x^{6} + 2 \,{\left (4 \, B a + 3 \, A b\right )} x^{3} - 3 \, A a\right )} \sqrt{b x^{3} + a}}{18 \, x^{3}}, \frac{3 \,{\left (2 \, B a + 3 \, A b\right )} \sqrt{-a} x^{3} \arctan \left (\frac{\sqrt{b x^{3} + a} \sqrt{-a}}{a}\right ) +{\left (2 \, B b x^{6} + 2 \,{\left (4 \, B a + 3 \, A b\right )} x^{3} - 3 \, A a\right )} \sqrt{b x^{3} + a}}{9 \, x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/2)*(B*x^3+A)/x^4,x, algorithm="fricas")

[Out]

[1/18*(3*(2*B*a + 3*A*b)*sqrt(a)*x^3*log((b*x^3 - 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) + 2*(2*B*b*x^6 + 2*(4*
B*a + 3*A*b)*x^3 - 3*A*a)*sqrt(b*x^3 + a))/x^3, 1/9*(3*(2*B*a + 3*A*b)*sqrt(-a)*x^3*arctan(sqrt(b*x^3 + a)*sqr
t(-a)/a) + (2*B*b*x^6 + 2*(4*B*a + 3*A*b)*x^3 - 3*A*a)*sqrt(b*x^3 + a))/x^3]

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Sympy [A]  time = 28.6349, size = 223, normalized size = 2.03 \begin{align*} - A \sqrt{a} b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{\frac{3}{2}}} \right )} - \frac{A a \sqrt{b} \sqrt{\frac{a}{b x^{3}} + 1}}{3 x^{\frac{3}{2}}} + \frac{2 A a \sqrt{b}}{3 x^{\frac{3}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} + \frac{2 A b^{\frac{3}{2}} x^{\frac{3}{2}}}{3 \sqrt{\frac{a}{b x^{3}} + 1}} - \frac{2 B a^{\frac{3}{2}} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{\frac{3}{2}}} \right )}}{3} + \frac{2 B a^{2}}{3 \sqrt{b} x^{\frac{3}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} + \frac{2 B a \sqrt{b} x^{\frac{3}{2}}}{3 \sqrt{\frac{a}{b x^{3}} + 1}} + B b \left (\begin{cases} \frac{\sqrt{a} x^{3}}{3} & \text{for}\: b = 0 \\\frac{2 \left (a + b x^{3}\right )^{\frac{3}{2}}}{9 b} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(3/2)*(B*x**3+A)/x**4,x)

[Out]

-A*sqrt(a)*b*asinh(sqrt(a)/(sqrt(b)*x**(3/2))) - A*a*sqrt(b)*sqrt(a/(b*x**3) + 1)/(3*x**(3/2)) + 2*A*a*sqrt(b)
/(3*x**(3/2)*sqrt(a/(b*x**3) + 1)) + 2*A*b**(3/2)*x**(3/2)/(3*sqrt(a/(b*x**3) + 1)) - 2*B*a**(3/2)*asinh(sqrt(
a)/(sqrt(b)*x**(3/2)))/3 + 2*B*a**2/(3*sqrt(b)*x**(3/2)*sqrt(a/(b*x**3) + 1)) + 2*B*a*sqrt(b)*x**(3/2)/(3*sqrt
(a/(b*x**3) + 1)) + B*b*Piecewise((sqrt(a)*x**3/3, Eq(b, 0)), (2*(a + b*x**3)**(3/2)/(9*b), True))

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Giac [A]  time = 1.22006, size = 139, normalized size = 1.26 \begin{align*} \frac{2 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} B b + 6 \, \sqrt{b x^{3} + a} B a b + 6 \, \sqrt{b x^{3} + a} A b^{2} + \frac{3 \,{\left (2 \, B a^{2} b + 3 \, A a b^{2}\right )} \arctan \left (\frac{\sqrt{b x^{3} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{3 \, \sqrt{b x^{3} + a} A a b}{x^{3}}}{9 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/2)*(B*x^3+A)/x^4,x, algorithm="giac")

[Out]

1/9*(2*(b*x^3 + a)^(3/2)*B*b + 6*sqrt(b*x^3 + a)*B*a*b + 6*sqrt(b*x^3 + a)*A*b^2 + 3*(2*B*a^2*b + 3*A*a*b^2)*a
rctan(sqrt(b*x^3 + a)/sqrt(-a))/sqrt(-a) - 3*sqrt(b*x^3 + a)*A*a*b/x^3)/b